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Sav [38]
3 years ago
13

50.00 ml of nitric acid (of unknown

Chemistry
1 answer:
mr Goodwill [35]3 years ago
5 0

Answer:

A)HNO_3 + NaOH → NaNO_3 + H2O

B) Explained below

C)volume of NaOH at the equivalence point will simply be the volume added to reach a pH value of 7.

D)the pH at the equivalence point is 7.

E)C(HNO3) = 0.1 × (volume of NaOH at equivalence point)/50

Explanation:

A) A symbolic equation for the reaction is;

HNO_3 + NaOH → NaNO_3 + H2O

B) HNO3 is a strong acid and NaOH is a strong base and as such they will both dissociate almost fully in a water solution to form ions. Thus, the reaction equation can now be rewritten in form of:

H(+) + OH(-) → H2O

From this reaction equation, it's clear that when sodium hydroxide (NaOH) was added, the positive hydrogen ions (H+) react with the negative hydroxyl ions (OH-) to form water (H2O).

Thus, the concentration [H+] will decrease because of both dilution and the decreasing amount of positive hydrogen ions (H+) in the solution. From definition, the pH is given by the formula;

pH = −log[H+]

Thus, pH increases as NaOH is added to HNO3 as given in the question.

C) The volume of NaOH at the equivalence point will simply be the volume added to reach a pH value of 7.

D) Since the given acid and base are very strong, it means that the solution at equivalence point is likely to be neutral. Thus, the pH at the equivalence point is 7.

E) Concentration of the nitric acid solution is given by;

C(HNO3) = (volume of NaOH at equivalence point × concentration of NaOH)/(volume of HNO3)

We are given;

concentration of NaOH) = 0.1 mol/L

Volume of HNO3 = 50 mL

Thus;

C(HNO3) = 0.1 × (volume of NaOH at equivalence point)/50

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Valentin [98]

Answer:

0.54 mole of H2O.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

2CH3OH + 3O2 —> 2CO2 + 4H2O

From the balanced equation above,

2 moles of CH3OH reacted to produce 4 moles of water.

Finally, we shall determine the number of mole of water (H2O) produced by the reaction of 0.27 moles of CH3OH. This can be obtained as follow:

From the balanced equation above,

2 moles of CH3OH reacted to produce 4 moles of water.

Therefore, 0.27 moles of CH3OH will react to produce = (0.27 × 4)/2 = 0.54 mole of H2O.

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3 years ago
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Which statement describes the light reaction of photosynthesis?
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If 125.0g of nitrogen is reacted with 125.0g of hydrogen, what is the theoretical yield of the reaction? What is the excess reac
MakcuM [25]

Answer:

Hydrogen is the excess reactant

Nitrogen is the limiting reactant

151.6g is theoretical yield

Explanation:

The reaction of N₂ with H₂ to produce NH₃ is:

N₂ + 3H₂ → 2NH₃

To find theoretical yield we need to determine limiting reactant with the moles of each gas as follows:

Nitrogen -Molar mass: 28g/mol-

125.0g * (1mol / 28g) = 4.46 moles

Hydrogen -Molar mass: 2g/mol-

125.0g * (1mol / 2g) = 62.5 moles of hydrogen

For a complete reaction of 4.46 moles of N2 there are needed:

4.46 moles N2 * (3moles H2 / 1mol N2) = 13.38 moles of hydrogen

As there are 62.5 moles of hydrogen:

<h3>Hydrogen is the excess reactant</h3><h3>Nitrogen is the limiting reactant</h3><h3 />

With nitrogen, the limiting reactant, we determine theoretical moles (Assuming 100% of the reaction occurs) and theoretical yield (In mass):

4.46 moles N2 * (2moles NH3 / 1mol N2) = 8.92 moles of ammonia

As molar mass of ammonia is 17g/mol:

8.92 moles of ammonia * (17g/mol) =

<h3>151.6g is theoretical yield</h3>

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Ilya [14]

Answer:

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Explanation:

1 molecule of CO_{2} contains 2 atoms of O

So, (6.023\times 10^{23}) molecules of  CO_{2} contains (2\times6.023\times10^{23}) atoms of O.

We know that 1 mol of an atom/molecule/ion represents 6.023\times10^{23} numbers of atoms/molecules/ions respectively.

So, (6.023\times 10^{23}) molecules of  CO_{2} is equal to 1 mol of CO_{2}.

(2\times6.023\times10^{23}) atoms of O is equal to 2 moles of O atom.

Hence, 1 mol of CO_{2} contains 2 moles of O atom.

Therefore, (7.9\times10^{-1}) mol of CO_{2} contains (2\times7.9\times10^{-1}) moles of O atom or 1.6 moles of O atom.

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